Programming in Java: Review of Class XI for ISC Class 12 Computer Science

What Programming in Java (Review of Class XI Sections B and C) is about

Weightage: CISCE publishes no chapter or unit weightage for ISC Computer Science. Programming in Java (Review of Class XI Sections B and C) is part of Section B, 4 of the 2027 syllabus and can be asked anywhere in the 70-mark Paper 1 (Part I compulsory: ten 1-mark items and a 10-mark short-answer question; Part II: two of three 10-mark questions in each of Sections A and B, and two of three 5-mark questions in Section C).Source: CISCE ISC Computer Science (868) syllabus, examination year 2027; latest Paper 1 specimen (2025) and the 2025 board paper

Key concepts

How Java runs: source, bytecode and the JVM

1. Source code is written in a .java file.
2. The compiler (javac) translates it into bytecode (.class file), an intermediate code that is not specific to any processor.
3. The Java Virtual Machine (JVM) on each computer interprets (and just-in-time compiles) the bytecode into machine instructions.

This gives platform independence ('write once, run anywhere'): the same .class file runs on any system with a JVM.

Other features of Java: object-oriented; robust (strong type checking, exception handling, automatic garbage collection); secure; supports multithreading; a large standard library (java.lang, java.util, java.io).

The JDK (Java Development Kit) contains the compiler and tools; the JRE (Java Runtime Environment) contains the JVM and libraries needed to run programs. BlueJ is an IDE that lets you create objects interactively and call methods without writing main( ), which is useful for learning, but board answers must still include main( ).

Worked example

[2] What is bytecode, and why does it make Java platform independent? (Understand)

Try it first, then show the solutionSolution

1. Bytecode: the intermediate code produced by the Java compiler in a .class file - 1 mark.
2. Any machine with a JVM can run the same bytecode, so the program need not be recompiled for each platform - 1 mark.

Answer: Bytecode is the machine-independent intermediate code the Java compiler produces. Because every platform has its own JVM that executes bytecode, the same compiled program runs on any system: Java is platform independent.

Structure of a Java class

A typical ISC program:

import java.util.Scanner;        // library import

public class Name                // class header
{
    int a;                        // data members (instance variables)
    static int count;             // class variable

    Name()                        // constructor
    {
        a = 0;
    }

    void accept() { }             // methods
    int compute(int x) { return x * 2; }

    public static void main(String[] args)
    {
        Name ob = new Name();     // create an object
        ob.accept();              // call methods
    }
}

Key points:
1. Only one public class per file, and the file name must match it.
2. main( ) is public, static, void and takes String[] args; execution starts there.
3. Statements end with semicolons; blocks are enclosed in braces.
4. Comments: // single line, /* multi-line */, /** documentation */.
5. Identifiers: letters, digits, _ and the currency sign; cannot start with a digit; cannot be a keyword; Java is case-sensitive.
6. Naming conventions: classes in PascalCase (KaprekarRange), methods and variables in camelCase (isKaprekar), constants in UPPER_CASE.

Worked example

[2] Why is main( ) declared static, and what is the role of String[] args? (Understand)

Try it first, then show the solutionSolution

1. static: the JVM can call main( ) without first creating an object of the class - 1 mark.
2. String[] args receives command-line arguments as an array of strings - 1 mark.

Answer: main( ) is static so that the JVM can call it using the class name before any object exists; String[] args holds any command-line arguments passed to the program.

Operators and output prediction

Arithmetic: + - * / % (integer division truncates: 7 / 2 = 3; % gives the remainder: 7 % 2 = 1).
Relational: < <= > >= == != (result boolean).
Logical: && (and), || (or), ! (not), with short-circuit evaluation.
Increment and decrement: x++ (use then increase), ++x (increase then use).
Assignment: = and compound += -= *= /= %=.
Ternary: condition ? a : b.

The + operator with strings: once a String operand appears (scanning left to right), + means concatenation. "ISC" + 1 + 2 gives ISC12, but 1 + 2 + "ISC" gives 3ISC.

Outputs of a review program:

// Short review fragments used on the page: operators, casting, conditional and loop outputs.
public class Review
{
    public static void main(String[] args)
    {
        int a = 7, b = 2;
        System.out.println(a / b + " " + a % b + " " + (double) a / b);
        int x = 5;
        int y = x++ + ++x;            // 5 + 7
        System.out.println(x + " " + y);
        char ch = 'A';
        ch += 2;
        System.out.println(ch + " " + (int) ch);
        String s = "ISC";
        System.out.println(s + 1 + 2);
        System.out.println(1 + 2 + s);
        int sum = 0;
        for (int i = 1; i <= 10; i += 3)
            sum += i;                 // 1 + 4 + 7 + 10
        System.out.println(sum);
        int k = 10;
        do
        {
            k -= 4;
        } while (k > 0);
        System.out.println(k);
    }
}

Output:

3 1 3.5
7 12
C 67
ISC12
3ISC
22
-2

Working: 7 / 2 = 3, 7 % 2 = 1, (double) 7 / 2 = 3.5; y = x++ + ++x = 5 + 7 = 12 and x ends at 7; 'A' + 2 = 'C' (code 67); the for loop adds 1, 4, 7, 10 = 22; the do-while subtracts 4 from 10 until k is not positive: 6, 2, -2.

Worked example

[2] Give the output: int x = 5; int y = x++ + ++x; System.out.println(x + " " + y); (Apply)

Try it first, then show the solutionSolution

1. x++ uses 5, then x becomes 6; ++x makes x 7 and uses 7 - 1 mark.
2. y = 5 + 7 = 12; prints 7 12 - 1 mark.

Answer: 7 12

Control flow review: decisions and loops

Decisions: if, if-else, if-else-if ladder, nested if, switch-case (with break to prevent fall-through; default for other values), ternary operator.
Loops:

• for: when the number of iterations is known (counter-controlled).
• while: entry-controlled; may run zero times.
• do-while: exit-controlled; runs at least once.

Jumps: break exits the nearest loop or switch; continue skips to the next iteration.

Nested loops are used for patterns and 2D arrays. Pascal's triangle shows a useful idea: each entry can be calculated from the previous one in the same row.

// Prints the first n rows of Pascal's triangle using nested loops.
// Each entry is found from the previous one: c = c * (r - k) / (k + 1).
public class Pascal
{
    public static void main(String[] args)
    {
        int n = 5;
        for (int r = 0; r < n; r++)
        {
            int c = 1;
            for (int k = 0; k <= r; k++)
            {
                System.out.print(c + " ");
                c = c * (r - k) / (k + 1);
            }
            System.out.println();
        }
    }
}

Output:

1 
1 1 
1 2 1 
1 3 3 1 
1 4 6 4 1

Worked example

[2] How many times does the body of do { k -= 4; } while (k > 0); run when k starts at 10, and what is the final k? (Apply)

Try it first, then show the solutionSolution

1. k: 10 → 6 (6 > 0, repeat) → 2 (repeat) → -2 (stop) - 1 mark.
2. The body runs 3 times; final k = -2 - 1 mark.

Answer: Three times; k ends at -2.

Higher-order problem solving: special numbers

ISC programs often test a number against a definition. Build them from reusable steps: count digits, extract digits, reverse, sum of digits, divisors, prime test, square and split.

Kaprekar number: a positive integer n with d digits such that when n² is split into a right part with d digits and a left part with the remaining digits, the two parts add up to n (the right part must not be 0). Examples: 9 (81 → 8 + 1), 45 (2025 → 20 + 25), 297 (88209 → 88 + 209).

import java.util.Scanner;

// Lists the Kaprekar numbers between p and q.
// n (with d digits) is a Kaprekar number if n*n split into a right part of d digits
// and a left part (the remaining digits) gives parts whose sum is n. Example: 45*45 = 2025, 20 + 25 = 45.
public class KaprekarRange
{
    int p, q;

    KaprekarRange()
    {
        p = 0;
        q = 0;
    }

    void accept()
    {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the lower and upper limits:");
        p = sc.nextInt();
        q = sc.nextInt();
    }

    // number of digits in n
    int digits(int n)
    {
        int c = 0;
        while (n > 0)
        {
            c++;
            n = n / 10;
        }
        return c;
    }

    boolean isKaprekar(int n)
    {
        long sq = (long) n * n;
        long div = 1;
        for (int i = 1; i <= digits(n); i++)
            div = div * 10;              // 10 raised to the number of digits
        long right = sq % div;
        long left = sq / div;
        return right > 0 && left + right == n;
    }

    void display()
    {
        int count = 0;
        System.out.print("Kaprekar numbers:");
        for (int n = p; n <= q; n++)
        {
            if (isKaprekar(n))
            {
                System.out.print(" " + n);
                count++;
            }
        }
        System.out.println();
        System.out.println("Frequency: " + count);
    }

    public static void main(String[] args)
    {
        KaprekarRange ob = new KaprekarRange();
        ob.accept();
        ob.display();
    }
}

Input: 1 1000

Enter the lower and upper limits:
Kaprekar numbers: 1 9 45 55 99 297 703 999
Frequency: 8

Points to note: n * n is stored in a long to avoid int overflow for larger n; div = 10^d is built with a loop (or (long) Math.pow(10, d)); the right part must be positive.

Worked example

[2] Show that 703 is a Kaprekar number. (Apply)

Try it first, then show the solutionSolution

1. 703² = 494209; 703 has 3 digits, so split as 494 | 209 - 1 mark.
2. 494 + 209 = 703, and the right part is not zero, so 703 is a Kaprekar number - 1 mark.

Answer: 703² = 494209 → 494 + 209 = 703, so 703 is a Kaprekar number.

Programming in Java (Review of Class XI Sections B and C) in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• How Java Runs and the Structure of a Java Class
• Operators, Control Flow and Number Programs in Java

Programming in Java (Review of Class XI Sections B and C): syntax and key facts

Integer division and remainder

a / b truncates; a % b remainder; a == (a / b) * b + a % b

7 / 2 = 3, 7 % 2 = 1.

Increment

x++ uses then adds; ++x adds then uses

x = 5: x++ + ++x = 5 + 7.

String + number

"ISC" + 1 + 2 = ISC12; 1 + 2 + "ISC" = 3ISC

Left to right.

Pascal entry

next = current × (r - k) / (k + 1)

Row r, position k.

Kaprekar test

n² = left × 10^d + right; left + right == n and right > 0

d = digits of n.

How the board asks Programming in Java (Review of Class XI Sections B and C)

The question patterns to expect, and the method that works for each.

Tricks to remember Programming in Java (Review of Class XI Sections B and C)

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Source → Bytecode → JVM

  'Some Bears Jump': Source, Bytecode, JVM.

  Use it when: How Java runs.

• Loop choice

  Know the count: for. Maybe zero times: while. At least once: do-while.

  Use it when: Choosing loops.

• String takes over

  Once a String joins a +, everything after it is glued, not added.

  Use it when: Concatenation outputs.

Programming in Java (Review of Class XI Sections B and C): board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 6-mark question split into 2 + 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Programming in Java (Review of Class XI Sections B and C)

The slips examiners see most often, and what to write instead.

• ✗Mistake: Forgetting integer division truncates

  Output questions often hinge on 7 / 2 being 3 in int arithmetic.

  Wrong working

  int avg = (7 + 8) / 2; // expecting 7.5

  Right working

  double avg = (7 + 8) / 2.0; // 7.5

  ✓Do this: Make one operand double (2.0 or a cast) before dividing when a fraction is needed.

  Usually costs: 1 mark.

• ✗Mistake: Missing main( ) or object creation

  CISCE examiners' comments, 2025: in Section B, 'Object creation and method invocation in main( ) wrong, or main( ) omitted; no documentation'. The same report asks teachers to 'Emphasise the importance of writing a main method and commenting the code properly'.

  ✓Do this: Finish every program with main( ) that creates the object and calls the methods; comment each method.

  Usually costs: 1-2 marks.

• ✗Mistake: Confusing break's role in switch and loops

  CISCE examiners' comments, 2025: on an assertion about break preventing fall-through, candidates 'Confused between options A and B because of the word loop in the Reason'.

  ✓Do this: break exits the nearest switch or loop; in a switch it prevents fall-through. Decide whether the Reason actually explains the Assertion.

  Usually costs: 1 mark.

• ✗Mistake: Overflow in squares and products

  int holds values up to about 2.1 × 10^9; squaring a 5-digit number can overflow.

  ✓Do this: Store large products in long: long sq = (long) n * n;

  Usually costs: Wrong output for large inputs.

Practice questions on Programming in Java (Review of Class XI Sections B and C)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · Very short answerBoard level

   What is the JVM?

   Hint

   Runs bytecode.

   Answer

   The Java Virtual Machine: software that executes Java bytecode on a particular computer.

2. Q2 · 1 mark · Very short answerBoard level

   Give the output of System.out.println(17 % 5 * 2);

   Hint

   Left to right, same precedence.

   Answer

   4 (17 % 5 = 2, then 2 * 2).

3. Q3 · 1 mark · Very short answerBoard level

   State one difference between while and do-while.

   Hint

   When is the condition tested?

   Answer

   while tests the condition before the body (may run zero times); do-while tests after (runs at least once).

4. Q4 · 1 mark · Very short answerHarder board level

   What is printed by int i = 0; while (i < 5) { if (i == 3) break; i++; } System.out.println(i);?

   Hint

   break when i is 3.

   Answer

   3

5. Q5 · 2 marks · Short answerHarder board level

   [2] Write a loop to print the first eight terms of 1, 1, 2, 3, 5, 8, ...

   Hint

   Keep two previous terms.

   Answer

   int a = 1, b = 1; for (int i = 1; i <= 8; i++) { System.out.print(a + " "); int c = a + b; a = b; b = c; } prints 1 1 2 3 5 8 13 21.

6. Q6 · 3 marks · Short answerHarder board level

   [3] Fill in ?1?-?3? so that the method returns the sum of the digits of n raised to the power of their position from the left (for 135: 1¹ + 3² + 5³ = 135). int disarium(int n) { String s = ?1?; int sum = 0; for (int i = 0; i < s.length(); i++) sum += (int) Math.pow(?2?, ?3?); return sum; }

   Hint

   Digit at i, power i + 1.

   Answer

   ?1? = String.valueOf(n) (or "" + n); ?2? = s.charAt(i) - '0'; ?3? = i + 1. For 135 the sum is 1 + 9 + 125 = 135.

7. Q7 · 5 marks · Long answerHOTS / competency

   [5] Write a method that prints all numbers from 100 to 999 that are equal to the sum of the cubes of their digits.

   Hint

   Armstrong numbers.

   Answer

   for (int n = 100; n <= 999; n++) { int t = n, s = 0; while (t > 0) { int d = t % 10; s += d * d * d; t /= 10; } if (s == n) System.out.print(n + " "); } prints 153 370 371 407.

8. Q8 · 10 marks · Long answerHOTS / competency

   [10] Design a class Fibo with data members n (number of terms) and an int array f; a constructor; void accept( ); void generate( ) that fills f with the first n Fibonacci terms (1, 1, 2, ...); and void showPrimes( ) that prints the terms that are prime. Write main( ).

   Hint

   Fill the array, then test each term.

   Answer

   Constructor sets n = 0; accept( ) reads n and creates f = new int[n]; generate( ) sets f[0] = f[1] = 1 (if n ≥ 2) and f[i] = f[i - 1] + f[i - 2]; showPrimes( ) prints f[i] if it has no divisor from 2 up to its square root and is at least 2; main( ) calls accept( ), generate( ), showPrimes( ). For n = 10 the terms are 1 1 2 3 5 8 13 21 34 55 and the primes printed are 2 3 5 13.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What does javac produce?Show answer

   Bytecode in a .class file.

2. 2.7 % 3?Show answer

   1.

3. 3.Loop that always runs at least once?Show answer

   do-while.

4. 4.1 + 2 + "A"?Show answer

   "3A".

5. 5.Kaprekar numbers up to 100?Show answer

   1, 9, 45, 55, 99.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Source (.java) → javac → bytecode (.class) → JVM.
• ☐main( ): public static void main(String[] args).
• ☐Integer division truncates; % gives remainder.
• ☐x++ uses then increments; ++x increments then uses.
• ☐String + anything = concatenation (left to right).
• ☐for (known count), while (0+ times), do-while (1+ times); break, continue.
• ☐Use long for large squares; build methods for digits, primes, reverse.

Programming in Java (Review of Class XI Sections B and C): frequently asked questions